Level 2 — RecallLinear Algebra (Full)

Linear Algebra (Full)

40 marksprintable — key stays hidden on paper

Level: 2 (Recall — definitions, standard problems, short derivations) Time: 30 minutes Total Marks: 40

Answer all questions. Show working. Use $...$ for math.

Q1. Let $\mathbf{u} = (1, 2, -2)$ and $\mathbf{v} = (2, -1, 2)$ . (a) Compute $\mathbf{u} \cdot \mathbf{v}$ . (1) (b) Compute $\|\mathbf{u}\|$ and $\|\mathbf{v}\|$ . (2) (c) Find the cosine of the angle between them. (2) (5 marks)

Q2. For $\mathbf{a} = (1, 0, 1)$ and $\mathbf{b} = (0, 1, 1)$ : (a) Compute the cross product $\mathbf{a} \times \mathbf{b}$ . (2) (b) State the geometric meaning of $\|\mathbf{a} \times \mathbf{b}\|$ and compute the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$ . (2) (4 marks)

Q3. Find the scalar projection and the vector projection of $\mathbf{u} = (3, 4)$ onto $\mathbf{v} = (1, 0)$ . (3 marks)

Q4. Solve the following system by Gaussian elimination: $x + y + z = 6,\quad 2x - y + z = 3,\quad x + 2y - z = 2.$ State the solution. (6 marks)

Q5. Let $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ , $B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ . (a) Compute $AB$ and $BA$ . (3) (b) State whether matrix multiplication is commutative, justifying from (a). (1) (4 marks)

Q6. For the matrix $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{pmatrix},$ find the rank by reducing to row echelon form. (4 marks)

Q7. Determine whether the vectors $(1, 0, 1)$ , $(0, 1, 1)$ , $(1, 1, 2)$ are linearly independent. Justify. (4 marks)

Q8. Compute the determinant of $A = \begin{pmatrix} 2 & 0 & 1 \\ 1 & 3 & 2 \\ 0 & 1 & 1 \end{pmatrix}$ by cofactor expansion along the first column. (4 marks)

Q9. Find the eigenvalues of $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ by solving the characteristic polynomial. (3 marks)

Q10. State the Rank–Nullity theorem for an $m \times n$ matrix $A$ , and use it to find the dimension of the null space of a $3 \times 5$ matrix of rank $2$ . (3 marks)

END OF PAPER

Answer keyMark scheme & solutions

Q1. (a) $\mathbf{u}\cdot\mathbf{v} = (1)(2)+(2)(-1)+(-2)(2) = 2-2-4 = -4$ . (1) (b) $\|\mathbf{u}\| = \sqrt{1+4+4} = \sqrt9 = 3$ ; $\|\mathbf{v}\| = \sqrt{4+1+4} = 3$ . (2, 1 each) (c) $\cos\theta = \dfrac{\mathbf{u}\cdot\mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|} = \dfrac{-4}{9}$ . (2: formula 1, value 1)

Q2. (a) $\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ 1&0&1\\0&1&1\end{vmatrix} = \mathbf i(0\cdot1 - 1\cdot1) - \mathbf j(1\cdot1 - 1\cdot0) + \mathbf k(1\cdot1 - 0\cdot0) = (-1, -1, 1)$ . (2) (b) $\|\mathbf a\times\mathbf b\|$ is the area of the parallelogram spanned by $\mathbf a,\mathbf b$ (1). Area $=\sqrt{1+1+1}=\sqrt3$ (1). (2)

Q3. Scalar projection $=\dfrac{\mathbf u\cdot\mathbf v}{\|\mathbf v\|} = \dfrac{3}{1} = 3$ (1.5). Vector projection $=\dfrac{\mathbf u\cdot\mathbf v}{\|\mathbf v\|^2}\mathbf v = 3(1,0) = (3,0)$ (1.5).

Q4. Augmented matrix; eliminate: $R_2\to R_2-2R_1: (0,-3,-1\,|\,-9)$ ; $R_3\to R_3-R_1: (0,1,-2\,|\,-4)$ . $R_3\to R_3 + \tfrac13 R_2: (0,0,-\tfrac73\,|\,-7)\Rightarrow z=3$ . (3 for elimination) Back-sub: $-3y - z = -9 \Rightarrow -3y = -6 \Rightarrow y=2$ ; $x+y+z=6\Rightarrow x=1$ . (3) Solution: $(x,y,z) = (1,2,3)$ .

Q5. (a) $AB = \begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix}2&1\\4&3\end{pmatrix}$ (1.5). $BA = \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&2\\3&4\end{pmatrix} = \begin{pmatrix}3&4\\1&2\end{pmatrix}$ (1.5). (3) (b) Not commutative: $AB\neq BA$ . (1)

Q6. $R_2\to R_2-2R_1 = (0,0,0)$ ; $R_3\to R_3-R_1=(0,-1,-2)$ . Rows: $(1,2,3),(0,-1,-2),(0,0,0)$ → 2 nonzero pivot rows. Rank $=2$ . (4: elimination 3, answer 1)

Q7. Form matrix with these as rows; row-reduce: $(1,0,1),(0,1,1),(1,1,2)$ . $R_3\to R_3-R_1=(0,1,1)$ , then $R_3\to R_3-R_2=(0,0,0)$ . A zero row appears → rank $2 < 3$ . Linearly dependent. Indeed $(1,1,2)=(1,0,1)+(0,1,1)$ . (4)

Q8. Expand along column 1: $\det A = 2\begin{vmatrix}3&2\\1&1\end{vmatrix} - 1\begin{vmatrix}0&1\\1&1\end{vmatrix} + 0$ $= 2(3-2) - 1(0-1) = 2(1) - 1(-1) = 2+1 = 3.$ (4: cofactors 2, minors 1, total 1)

Q9. $\det(A-\lambda I) = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda-1)(\lambda-3)$ . Eigenvalues $\lambda = 1, 3$ . (3: char poly 2, roots 1)

Q10. Rank–Nullity: $\operatorname{rank}(A) + \dim(\operatorname{Null}(A)) = n$ (number of columns). (2) For $3\times5$ , $n=5$ , rank $=2$ : nullity $=5-2=3$ . (1)

[
  {"claim":"u·v = -4 and cos = -4/9", "code":"u=Matrix([1,2,-2]); v=Matrix([2,-1,2]); d=(u.T*v)[0]; c=d/(u.norm()*v.norm()); result = (d==-4) and (c==Rational(-4,9))"},
  {"claim":"cross product (1,0,1)x(0,1,1)=(-1,-1,1), area sqrt3", "code":"a=Matrix([1,0,1]); b=Matrix([0,1,1]); cr=a.cross(b); result = (cr==Matrix([-1,-1,1])) and (cr.norm()==sqrt(3))"},
  {"claim":"system solution (1,2,3)", "code":"x,y,z=symbols('x y z'); sol=solve([x+y+z-6,2*x-y+z-3,x+2*y-z-2],[x,y,z]); result = sol=={x:1,y:2,z:3}"},
  {"claim":"det = 3", "code":"A=Matrix([[2,0,1],[1,3,2],[0,1,1]]); result = A.det()==3"},
  {"claim":"eigenvalues 1 and 3", "code":"A=Matrix([[2,1],[1,2]]); result = set(A.eigenvals().keys())=={1,3}"}
]